Extending a square matrix to a unitary matrix
$begingroup$
Suppose we have a square matrix $M$ of size $ntimes n$. It is given that any element $M_{ij}$ of $M$ is a real number and satisfies $0 leq M_{ij} leq 1$, $forall$ $i,j$. No other property for $M$ is known. Is it possible to create a new matrix $U$, $s.t.$:
$U$ is a square matrix of size $2ntimes 2n$,
$U$ is of the form $begin{bmatrix}M&A\B&Cend{bmatrix}$,
$A,B,C$ are all of size $ntimes n$ and all of $A,B,C$ are unique linear transformations of $M$,- The elements of $A,B,C$ can take complex values,
- And that $U$ is unitary, $i.e.$, $UU^dagger = U^dagger U = I$, (where $I$ is the identitiy matrix, and $U^dagger$ is symbol for complex conjugate of $U$)?
Thank you
quantum-gate mathematics unitarity
edited Jan 14 at 16:55
Blue♦
5,90121354
asked Jan 10 at 10:26
new2quantumnew2quantum
211
$endgroup$
add a comment |
$begingroup$
Suppose we have a square matrix $M$ of size $ntimes n$. It is given that any element $M_{ij}$ of $M$ is a real number and satisfies $0 leq M_{ij} leq 1$, $forall$ $i,j$. No other property for $M$ is known. Is it possible to create a new matrix $U$, $s.t.$:
$U$ is a square matrix of size $2ntimes 2n$,
$U$ is of the form $begin{bmatrix}M&A\B&Cend{bmatrix}$,
$A,B,C$ are all of size $ntimes n$ and all of $A,B,C$ are unique linear transformations of $M$,- The elements of $A,B,C$ can take complex values,
- And that $U$ is unitary, $i.e.$, $UU^dagger = U^dagger U = I$, (where $I$ is the identitiy matrix, and $U^dagger$ is symbol for complex conjugate of $U$)?
Thank you
quantum-gate mathematics unitarity
edited Jan 14 at 16:55
Blue♦
5,90121354
asked Jan 10 at 10:26
new2quantumnew2quantum
211
$endgroup$
$begingroup$
To clarify for 3, do you mean along the lines of "there exist P and Q such that PMQ=A"? Similarly for B,C?
$endgroup$
– AHusain
Jan 10 at 10:34
$begingroup$
actually more relaxed. what i mean is that $A,B,C$ may be obtained by some transformation (perhaps not linear, not sure) of $M$. for instance $A,B,C$ may be $-M$. What would also work is that $A,B,C$ may be for instance $frac{1}{sqrt{n}}$ times the Identity and so on. Basically, we are given $M$ and need to create $2n times 2n$ unitary matrix from it.
$endgroup$
– new2quantum
Jan 10 at 10:59
add a comment |
$begingroup$
Suppose we have a square matrix $M$ of size $ntimes n$. It is given that any element $M_{ij}$ of $M$ is a real number and satisfies $0 leq M_{ij} leq 1$, $forall$ $i,j$. No other property for $M$ is known. Is it possible to create a new matrix $U$, $s.t.$:
$U$ is a square matrix of size $2ntimes 2n$,
$U$ is of the form $begin{bmatrix}M&A\B&Cend{bmatrix}$,
$A,B,C$ are all of size $ntimes n$ and all of $A,B,C$ are unique linear transformations of $M$,- The elements of $A,B,C$ can take complex values,
- And that $U$ is unitary, $i.e.$, $UU^dagger = U^dagger U = I$, (where $I$ is the identitiy matrix, and $U^dagger$ is symbol for complex conjugate of $U$)?
Thank you
quantum-gate mathematics unitarity
edited Jan 14 at 16:55
Blue♦
5,90121354
asked Jan 10 at 10:26
new2quantumnew2quantum
211
$endgroup$
Suppose we have a square matrix $M$ of size $ntimes n$. It is given that any element $M_{ij}$ of $M$ is a real number and satisfies $0 leq M_{ij} leq 1$, $forall$ $i,j$. No other property for $M$ is known. Is it possible to create a new matrix $U$, $s.t.$:
$U$ is a square matrix of size $2ntimes 2n$,
$U$ is of the form $begin{bmatrix}M&A\B&Cend{bmatrix}$,
$A,B,C$ are all of size $ntimes n$ and all of $A,B,C$ are unique linear transformations of $M$,- The elements of $A,B,C$ can take complex values,
- And that $U$ is unitary, $i.e.$, $UU^dagger = U^dagger U = I$, (where $I$ is the identitiy matrix, and $U^dagger$ is symbol for complex conjugate of $U$)?
Thank you
quantum-gate mathematics unitarity
quantum-gate mathematics unitarity
edited Jan 14 at 16:55
Blue♦
5,90121354
asked Jan 10 at 10:26
new2quantumnew2quantum
211
edited Jan 14 at 16:55
Blue♦
5,90121354
asked Jan 10 at 10:26
new2quantumnew2quantum
211
edited Jan 14 at 16:55
Blue♦
5,90121354
edited Jan 14 at 16:55
Blue♦
5,90121354
edited Jan 14 at 16:55
Blue♦
5,90121354
5,90121354
asked Jan 10 at 10:26
new2quantumnew2quantum
211
asked Jan 10 at 10:26
new2quantumnew2quantum
211
asked Jan 10 at 10:26
new2quantumnew2quantum
211
211
$begingroup$
To clarify for 3, do you mean along the lines of "there exist P and Q such that PMQ=A"? Similarly for B,C?
$endgroup$
– AHusain
Jan 10 at 10:34
$begingroup$
actually more relaxed. what i mean is that $A,B,C$ may be obtained by some transformation (perhaps not linear, not sure) of $M$. for instance $A,B,C$ may be $-M$. What would also work is that $A,B,C$ may be for instance $frac{1}{sqrt{n}}$ times the Identity and so on. Basically, we are given $M$ and need to create $2n times 2n$ unitary matrix from it.
$endgroup$
– new2quantum
Jan 10 at 10:59
add a comment |
$begingroup$
To clarify for 3, do you mean along the lines of "there exist P and Q such that PMQ=A"? Similarly for B,C?
$endgroup$
– AHusain
Jan 10 at 10:34
$begingroup$
actually more relaxed. what i mean is that $A,B,C$ may be obtained by some transformation (perhaps not linear, not sure) of $M$. for instance $A,B,C$ may be $-M$. What would also work is that $A,B,C$ may be for instance $frac{1}{sqrt{n}}$ times the Identity and so on. Basically, we are given $M$ and need to create $2n times 2n$ unitary matrix from it.
$endgroup$
– new2quantum
Jan 10 at 10:59
$begingroup$
To clarify for 3, do you mean along the lines of "there exist P and Q such that PMQ=A"? Similarly for B,C?
$endgroup$
– AHusain
Jan 10 at 10:34
$begingroup$
To clarify for 3, do you mean along the lines of "there exist P and Q such that PMQ=A"? Similarly for B,C?
$endgroup$
– AHusain
Jan 10 at 10:34
$begingroup$
actually more relaxed. what i mean is that $A,B,C$ may be obtained by some transformation (perhaps not linear, not sure) of $M$. for instance $A,B,C$ may be $-M$. What would also work is that $A,B,C$ may be for instance $frac{1}{sqrt{n}}$ times the Identity and so on. Basically, we are given $M$ and need to create $2n times 2n$ unitary matrix from it.
$endgroup$
– new2quantum
Jan 10 at 10:59
$begingroup$
actually more relaxed. what i mean is that $A,B,C$ may be obtained by some transformation (perhaps not linear, not sure) of $M$. for instance $A,B,C$ may be $-M$. What would also work is that $A,B,C$ may be for instance $frac{1}{sqrt{n}}$ times the Identity and so on. Basically, we are given $M$ and need to create $2n times 2n$ unitary matrix from it.
$endgroup$
– new2quantum
Jan 10 at 10:59
add a comment |
1 Answer
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votes
$begingroup$
No. The rows and columns of a unitary $U$ must have a sum-mod-square of 1.
$$
sum_{i}|U_{ij}|^2=sum_{j}|U_{ij}|^2=1
$$
Your $M$, as specified, could have a 1 element along a whole row so the sum-mod-square of the corresponding row in $U$ would be $n$. So, unless $n=1$, it's impossible without further constraints on $M$.
answered Jan 10 at 10:37
DaftWullieDaftWullie
12.9k1539
$endgroup$
$begingroup$
i understand, so lets assume that $M$ can be normalized to any bound, to satisfy the condition you mention.
$endgroup$
– new2quantum
Jan 10 at 10:53
2
$begingroup$
See this answer.
$endgroup$
– DaftWullie
Jan 10 at 14:21
add a comment |
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1 Answer
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$begingroup$
No. The rows and columns of a unitary $U$ must have a sum-mod-square of 1.
$$
sum_{i}|U_{ij}|^2=sum_{j}|U_{ij}|^2=1
$$
Your $M$, as specified, could have a 1 element along a whole row so the sum-mod-square of the corresponding row in $U$ would be $n$. So, unless $n=1$, it's impossible without further constraints on $M$.
answered Jan 10 at 10:37
DaftWullieDaftWullie
12.9k1539
$endgroup$
$begingroup$
i understand, so lets assume that $M$ can be normalized to any bound, to satisfy the condition you mention.
$endgroup$
– new2quantum
Jan 10 at 10:53
2
$begingroup$
See this answer.
$endgroup$
– DaftWullie
Jan 10 at 14:21
add a comment |
$begingroup$
No. The rows and columns of a unitary $U$ must have a sum-mod-square of 1.
$$
sum_{i}|U_{ij}|^2=sum_{j}|U_{ij}|^2=1
$$
Your $M$, as specified, could have a 1 element along a whole row so the sum-mod-square of the corresponding row in $U$ would be $n$. So, unless $n=1$, it's impossible without further constraints on $M$.
answered Jan 10 at 10:37
DaftWullieDaftWullie
12.9k1539
$endgroup$
$begingroup$
i understand, so lets assume that $M$ can be normalized to any bound, to satisfy the condition you mention.
$endgroup$
– new2quantum
Jan 10 at 10:53
2
$begingroup$
See this answer.
$endgroup$
– DaftWullie
Jan 10 at 14:21
add a comment |
$begingroup$
No. The rows and columns of a unitary $U$ must have a sum-mod-square of 1.
$$
sum_{i}|U_{ij}|^2=sum_{j}|U_{ij}|^2=1
$$
Your $M$, as specified, could have a 1 element along a whole row so the sum-mod-square of the corresponding row in $U$ would be $n$. So, unless $n=1$, it's impossible without further constraints on $M$.
answered Jan 10 at 10:37
DaftWullieDaftWullie
12.9k1539
$endgroup$
No. The rows and columns of a unitary $U$ must have a sum-mod-square of 1.
$$
sum_{i}|U_{ij}|^2=sum_{j}|U_{ij}|^2=1
$$
Your $M$, as specified, could have a 1 element along a whole row so the sum-mod-square of the corresponding row in $U$ would be $n$. So, unless $n=1$, it's impossible without further constraints on $M$.
answered Jan 10 at 10:37
DaftWullieDaftWullie
12.9k1539
answered Jan 10 at 10:37
DaftWullieDaftWullie
12.9k1539
answered Jan 10 at 10:37
DaftWullieDaftWullie
12.9k1539
answered Jan 10 at 10:37
DaftWullieDaftWullie
12.9k1539
12.9k1539
$begingroup$
i understand, so lets assume that $M$ can be normalized to any bound, to satisfy the condition you mention.
$endgroup$
– new2quantum
Jan 10 at 10:53
2
$begingroup$
See this answer.
$endgroup$
– DaftWullie
Jan 10 at 14:21
add a comment |
$begingroup$
i understand, so lets assume that $M$ can be normalized to any bound, to satisfy the condition you mention.
$endgroup$
– new2quantum
Jan 10 at 10:53
2
$begingroup$
See this answer.
$endgroup$
– DaftWullie
Jan 10 at 14:21
$begingroup$
i understand, so lets assume that $M$ can be normalized to any bound, to satisfy the condition you mention.
$endgroup$
– new2quantum
Jan 10 at 10:53
$begingroup$
i understand, so lets assume that $M$ can be normalized to any bound, to satisfy the condition you mention.
$endgroup$
– new2quantum
Jan 10 at 10:53
2
2
$begingroup$
See this answer.
$endgroup$
– DaftWullie
Jan 10 at 14:21
$begingroup$
See this answer.
$endgroup$
– DaftWullie
Jan 10 at 14:21
add a comment |
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$begingroup$
To clarify for 3, do you mean along the lines of "there exist P and Q such that PMQ=A"? Similarly for B,C?
$endgroup$
– AHusain
Jan 10 at 10:34
$begingroup$
actually more relaxed. what i mean is that $A,B,C$ may be obtained by some transformation (perhaps not linear, not sure) of $M$. for instance $A,B,C$ may be $-M$. What would also work is that $A,B,C$ may be for instance $frac{1}{sqrt{n}}$ times the Identity and so on. Basically, we are given $M$ and need to create $2n times 2n$ unitary matrix from it.
$endgroup$
– new2quantum
Jan 10 at 10:59